lim_(n->oo)(1+1/n)^n = ?

3 Answers
Sep 22, 2016

Let a_n and b_n, n\inN, are defined as:

a_n=(1+1/n)^n,

b_n=(1+1/n)^(n+1) = a_n(1+1/n).

It is obvious that:

lim_(n->infty) a_n = (lim_(n->infty) b_n)/(lim_(n->infty) (1+1/n)) = lim_(n->infty) b_n.

We'll show that sequence b_n is decreasing.

b_(n-1)/b_n = (1+1/(n-1))^n/(1+1/n)^(n+1) = (n/(n-1))^n/((n+1)/n)^(n+1)=n^(2n+1)/((n-1)^n(n+1)^(n+1))

b_(n-1)/b_n = (n^2/(n^2-1))^n*n/(n+1) = (1+1/(n^2-1))^n*n/(n+1)

Using Bernoulli's inequality (1+x)^n > 1+nx it follows:

(1+1/(n^2-1))^n > 1+n/(n^2-1) and:

b_(n-1)/b_n > (1+n/(n^2-1))*n/(n+1) > (1+n/n^2)*n/(n+1)

b_(n-1)/b_n > (n(n+1))/n^2*n/(n+1) = 1

So, the sequence b_n is decreasing.

Using Bernoulli's inequality again:

b_n = (1+1/n)^(n+1) > 1+(n+1)/n = 1+1+1/n = 2+1/n > 2

it follows that b_n has a lower bound. b_n is convergent.

Using inequality of arithmetic and geometric means:

root(n+1)((1+1/n)^n*1) < (n(1+1/n)+1)/(n+1) = (n+2)/(n+1)

it follows:

(1+1/n)^n < ((n+2)/(n+1))^(n+1) = (1+1/(n+1))^(n+1),

a_n < a_(n+1)

and a_n is increasing.

a_n = (1+1/n)^n < (1+1/n)^(n+1) = b_n

It follows that a_n has an upper bound (because b_n is decreasing and has a lower bound). Sequence a_n is increasing and has an upper bound so it's convergent.

Finally,

e = lim_(n->infty) (1+1/n)^n.

Let some value x = lim_(n->oo) (1 + 1/n)^n, where n is an arbitrary variable (i.e. does not have to be an integer!). Then the fastest way to proceed is as follows.

lnx = ln(lim_(n->oo) (1+1/n)^n)

lnx = lim_(n->oo)(nln(1+1/n))

lnx = lim_(n->oo)(ln(1+1/n)/(1/n))

By inspection, we've converted the right side into a 0/0 form. Therefore, we can use L'Hopital's rule.

lnx = lim_(n->oo)((1/(1+1/n)*cancel(-1/n^2))/cancel(-1/n^2))

lnx = lim_(n->oo)(1/(1+1/n))

wherein the 1/n term vanishes as n->oo.

lnx = lim_(nrarroo)(1/(1+0))

lnx=lim_(nrarroo)1

lnx=1

Thus, we undo the natural logarithm to get:

color(blue)(lim_(n->oo) (1+1/n)^n)

= e^(ln(lim_(n->oo) (1+1/n)^n))

= e^(lnx) = x = e^1 = color(blue)(e).

Sep 22, 2016

(1+1/n)^n < (1+1/(n+1))^(n+1) is increasing for n->oo

(1+1/n)^n = 1+1/(1!)n/n+1/(2!)(n/n)((n-1)/n) + 1/(3!)(n/n)((n-1)/n)((n-2)/n)+cdots

((n-1)/n)(1+1/n)^n le 1-1+ (n-1)/n+1/(1!)(n-1)/n+1/(2!)((n-1)/n)^2+1/(3!)((n-1)/n)^3 + 1/(4!)((n-1)/n)^4+cdots +

((n-1)/n)(1+1/n)^n le (n-1)/n-1+sum_(k=0)^n1/(k!)((n-1)/n)^k

so

lim_(n->oo)((n-1)/n)(1+1/n)^n le lim_(n->oo)(-1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)((n-1)/n)^k lelim_(n->oo)( -1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)

Finally

lim_(n->oo)((n-1)/n)(1+1/n)^n = lim_(n->oo)(1+1/n)^n

and

lim_(n->oo)(1+1/n)^n le e